If a baseball is projected upward from ground level with an initial velocity of 128 feet per second, then its height is a function of time, given by s=-16t^2+128t.
The maximum height reached by the ball is _____ feet
2007-06-28 10:26:25 · 5 answers · asked by hllywdbabe7 1 in Science & Mathematics ➔ Mathematics
first you have to find t1 the time it took for the baseball to reach its maximum height. At its maximum height the speed of the baseball is 0. The speed is the derivative of the distance s so v=-32t+128 that means t=-(v-128)/32 and if v=0 then t=4
that means that s=(-16*16)+(128*4)=256ft
s=256ft
2007-06-28 10:48:28 · answer #1 · answered by ntsoaE 2 · 0⤊ 0⤋
s = - 16t² + 128t
ds/dt = - 32t + 128 = 0 at max. height
32t = 128
t = 128 / 32 = 32 / 8 = 4
s(4) = - 16 x 4² + 128 x 4
s(4) = - 256 + 512
s(4) = 256
The maximum height reached by the ball is 256 ft
2007-07-02 12:15:23 · answer #2 · answered by Como 7 · 0⤊ 0⤋
s=-16t^2+128t.
v=ds/dt=-32t+128
at the highestpoint, v=0
so -32t+128=0,t=4s
max height=-16x16+128x4=512-256=256 ft. answer
2007-06-28 18:19:18 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Well, I haven't done this kind of math for a while, but I think it's somewhere around 255.54. I may be completely wrong, though...
2007-06-28 17:34:36 · answer #4 · answered by gecko 2 · 0⤊ 0⤋
v = u + at
u = 128
v = 0
a = (-9.8 ms^-2)
t = ?
stick this value of t into equation for s
2007-06-28 17:31:30 · answer #5 · answered by Anonymous · 0⤊ 0⤋